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          <h1 class="post-title" itemprop="name headline">PAT B1024/A1073</h1>
        

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        <p>科学计数法是科学家用来表示很大或很小的数字的一种方便的方法，其满足正则表达式[+-][1-9]”.”[0-9]+E[+-][0-9]+，即数字的整数部分只有1位，小数部分至少有1位，该数字及其指数部分的正负号即使对正数也必定明确给出。</p>
<p>现以科学计数法的格式给出实数A，请编写程序按普通数字表示法输出A，并保证所有有效位都被保留。</p>
<p>输入格式：</p>
<p>每个输入包含1个测试用例，即一个以科学计数法表示的实数A。该数字的存储长度不超过9999字节，且其指数的绝对值不超过9999。</p>
<p>输出格式：</p>
<p>对每个测试用例，在一行中按普通数字表示法输出A，并保证所有有效位都被保留，包括末尾的0。</p>
<p>输入样例1：<br>+1.23400E-03<br>输出样例1：<br>0.00123400<br>输入样例2：<br>-1.2E+10<br>输出样例2：<br>-12000000000</p>
<p>Scientific notation is the way that scientists easily handle very large numbers or very small numbers. The notation matches the regular expression [+-][1-9]”.”[0-9]+E[+-][0-9]+ which means that the integer portion has exactly one digit, there is at least one digit in the fractional portion, and the number and its exponent’s signs are always provided even when they are positive.</p>
<p>Now given a real number A in scientific notation, you are supposed to print A in the conventional notation while keeping all the significant figures.</p>
<p>Input Specification:</p>
<p>Each input file contains one test case. For each case, there is one line containing the real number A in scientific notation. The number is no more than 9999 bytes in length and the exponent’s absolute value is no more than 9999.</p>
<p>Output Specification:</p>
<p>For each test case, print in one line the input number A in the conventional notation, with all the significant figures kept, including trailing zeros,</p>
<p>Sample Input 1:<br>+1.23400E-03<br>Sample Output 1:<br>0.00123400<br>Sample Input 2:<br>-1.2E+10<br>Sample Output 2:<br>-12000000000</p>
<pre><code>#include &quot;stdio.h&quot;
//#include &quot;math.h&quot;
#include &quot;string.h&quot;
//#include &quot;algorithm&quot;
//using namespace std;
int main(){
    char str[10010];
    gets(str);
    int len =  strlen(str);
    //输出符号
    if(str[0] == &apos;-&apos;)printf(&quot;-&quot;);
    //定位E的位置
    int posE = 0;
    while (str[posE] != &apos;E&apos;) posE++;

    int exp = 0;
    for (int i = posE + 2; i &lt; len ; i++)
        exp = exp * 10 + str[i] - &apos;0&apos;;
    if (exp == 0) {//指数为零
        for (int i = 1; i&lt; posE ; i++) {
            printf(&quot;%c&quot;, str[i]);
        }
    }
    //指数为负
    if (str[posE + 1] == &apos;-&apos;) {
        printf(&quot;0.&quot;);
        for (int i = 0; i &lt; exp - 1 ; i++) {
            printf(&quot;0&quot;);
        }
        printf(&quot;%c&quot;, str[1]);//小数点以外的数字
        for (int i = 3; i &lt; posE; i++) {
            printf(&quot;%c&quot;,str[i]);
        }
    }else {//指数为正
        for (int i = 1; i &lt; posE; i++) {
            if (str[i] == &apos;.&apos;) continue;
            printf(&quot;%c&quot;, str[i]);
            if (i == exp + 2 &amp;&amp; posE - 3 != exp) {
                printf(&quot;.&quot;);//如果E与小数点之间的距离不等于指数，就在exp+2的位置输出小数点
            }
        }
        for (int i = 0; i &lt; exp - (posE - 3); i++) {
            printf(&quot;0&quot;);
        }
    }
    return 0;
}
</code></pre>
      
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